So the question is really hard I think. I tried using a simple way by calculating the probability of each combination that makes a sum divisible by six, but it would take forever. Does anyone have any ideas?

Suppose that we roll a six-sided die ten times. What is the probability that the total of all ten rolls is divisible by six?

  • $\begingroup$ Almost a duplicate of this question. Using the same logic, the answer is easily seen to be $1/6$. $\endgroup$
    – TonyK
    Nov 11 '15 at 15:33


Roll $9$ times and let $x$ be the total.

For exactly one number $n\in\{1,2,3,4,5,6\}$ we will have $6 \mid (x+n)$ (i.e. $x+n$ is divisible by $6$).

  • 4
    $\begingroup$ I didn't get it, why would I roll 9 times instead of 10? and what is 6|x? $\endgroup$
    – Xlyon
    Nov 11 '15 at 9:38
  • 10
    $\begingroup$ @Xlyon When you've rolled nine times there's always exactly one outcome for the tenth that will make the sum divisible by $6$. $\endgroup$
    – skyking
    Nov 11 '15 at 9:43
  • 1
    $\begingroup$ Thanks! that is logical. Can you please help me complete it? $\endgroup$
    – Xlyon
    Nov 11 '15 at 10:50
  • 4
    $\begingroup$ Exactly one of the faces of the die thrown at the $10$th time will result in a total sum ($x+n$) that is divisible by $6$. The probability that this face shows up is $\frac16$ (if the die is fair, of course). $\endgroup$
    – drhab
    Nov 11 '15 at 10:54
  • 4
    $\begingroup$ The value of $x$ is indeed irrelevant. Whatever value $x$ takes, it's chance to change into a number divisible by $6$ (when the last result is added by $x$) is in all cases $\frac16$. So that's indeed the answer to this question. They could have asked: take some arbitrary number $y\in\mathbb Z$ and trhrow a fair die. If $D$ denotes the outcome then what is the probability that $y+D$ is divisible by $6$? Same answer: $\frac16$, $\endgroup$
    – drhab
    Nov 11 '15 at 11:34

After rolling the die once, there is equal probability for each result modulo 6. Adding any unrelated integer to it will preserve the equidistribution. So you can even roll a 20-sided die afterwards and add its outcome: the total sum will still have a probability of 1/6 to be divisible by 6.

  • 6
    $\begingroup$ @Kevin The d20 by itself does have that unequal distribution, but the sum d20+d6 does not. It literally does not matter what _**x**_ is in determining the distribution of (_**x**_ + d6) mod 6. In fact, d20 + d6 also has exactly a 1/20 chance of being divisible by 20 by the same logic $\endgroup$ Nov 11 '15 at 19:28

If you want something a little more formal and solid than drhab's clever and brilliant answer:

Let $P(k,n)$ be the probability of rolling a total with remainder $k$ when divided by $6, (k = 0...5)$ with $n$ die.

$P(k, 1)$ = Probability of rolling a $k$ if $k \ne 0$ or a $6$ if $k = 6$; $P(k, 1) = \frac 1 6$.

For $n > 1$. $ P(k,n) = \sum_{k= 0}^5 P(k, n-1)\cdot \text{Probability of Rolling(6-k)} = \sum_{k= 0}^5 P(k, n-1)\cdot\frac 1 6= \frac 1 6\sum_{k= 0}^5 P(k, n-1)= \frac 1 6 \cdot 1 = \frac 1 6$

This is drhab's answer but in formal terms without appeals to common sense

  • 2
    $\begingroup$ That's a way to do it. What I actually had in mind was: $P\left(6\text{ divides }S_{10}\right)=\sum_{m\in\mathbb{N}}P\left(6\text{ divides }S_{10}\mid S_{9}=m\right)P\left(S_{9}=m\right)=\sum_{m\in\mathbb{N}}\frac{1}{6}P\left(S_{9}=m\right)=\frac{1}{6}$ $\endgroup$
    – drhab
    Nov 12 '15 at 9:40
  • $\begingroup$ Yes, that is more direct. $\endgroup$
    – fleablood
    Nov 12 '15 at 16:45
  • $\begingroup$ This should be the accepted answer. It's formulated in accessible notation and actually makes sense. I did not trust the "common sense" argument in the answer of @drhab (and their comment to this answer is opaque to me). $\endgroup$
    – Kagaratsch
    Mar 1 at 21:00
  • 1
    $\begingroup$ I dunno. If you can simplify something done to an easy, almost trivial case, you should. A $\sum_{k=1}^{10} a_k \equiv 0 \pmod 6 \iff \sum_{k=1}^{9}a_k \equiv - a_{10}\pmod 6$. And for any value $M = \sum_{k=1}^{9}a_k$ we know the probablilty that $a_{10}\equiv M \pmod 6$ is exactly $\frac 16$. ... Now if I teaching probability I don't want students shying away from rolling up the sleeves and getting dirty but but If I want to do an answer I want things as simple as possible. If it doesn't matter what the first nine rolls are so long as the 10th cancels them... then why bother. $\endgroup$
    – fleablood
    Mar 1 at 21:35

In spite of all great answers, given here, I say, why not give another proof, from another point of view. The problem is we have 10 random variables $X_i$ for $i=1,\dots,10$, defined over $[6]=\{1,\dots,6\}$, and we are interested in distribution of $Z$ defined as $$ Z=X_1\oplus X_2\oplus \dots \oplus X_{10} $$ where $\oplus$ is addition modulo $6$. We can go on by two different, yet similar proofs.

First proof: If $X_1$ and $X_2$ are two random variables over $[6]$, and $X_1$ is uniformly distributed, sheer calculation can show that $X_1\oplus X_2$ is also uniformly distributed. Same logic yields that $Z$ is uniformly distributed over $[6]$.

Remark: This proves a more general problem. It says that even if only one of the dices is fair dice, i.e. each side appearing with probability $\frac 16$, the distribution of $Z$ will be uniform and hence $\mathbb P(Z=0)=\frac 16$.

Second proof: This proof draws on (simple) information theoretic tools and assumes its background. The random variable $Z$ is output of an additive noisy channel and it is known that the worst case is uniformly distributed noise. In other word if $X_i$ is uniform for only one $i$, $Z$ will be uniform. To see this, suppose that $X_1$ is uniformly distributed. Then consider the following mutual information $I(X_2,X_3,\dots,X_6;Z)$ which can be written as $H(Z)-H(Z|X_2,\dots,X_6)$. But we have: $$ H(Z|X_2,\dots,X_6)=H(X_1|X_2,\dots,X_6)=H(X_1) $$
where the first equality is resulted from the fact that knowing $X_2,\dots,X_6$ the only uncertainty in $Z$ is due to $X_1$. The second equality is because $X_1$ is independent of others. Know see that:

  • Mutual information is positive: $H(Z)\geq H(X_1)$
  • Entropy of $Z$ is always less that or equal to the entropy of uniformly distributed random variable over $[6]$: $H(Z)\leq H(X_1)$
  • From the last two $H(Z)=H(X_1)$ and $Z$ is uniformly distributed and the proof is complete.

Similarly here, only one fair dice is enough. Moreover the same proof can be used for an arbitrary set $[n]$. As long as one of the $X_i$'s is uniform, then their finite sum modulo $n$ will be uniformly distributed.


There are 3 variables in this case:

  • the number of sides of the dice: s (e.g. 6)
  • the number of throws: t (e.g. 10)
  • the requesed multiple: x (e.g. 6)

In this case, the conditions are simple:

  • s>=x
  • x >0
  • t > 0

And also the answer is simple: Throwing a sum that is a multiple of 6 has a 1/6 probability.

$P(s,t,x) = 1/x$

For situations where s<x this is not entirely correct. It approaches the same result though, at a high amount of throws. Example: If you throw a 6-sided dice 30 times the chance that the sum is a multiple of 20 will be about 5%. Proving this is a bit of a challenge.

$\lim \limits_{t \to \infty} P(s,t,x) = 1/x$,

Nevertheless, if programming is an acceptable proof:

public static void main(String[] args) {
    int t_throws = 10;
    int s_sides = 6;
    int x_multiple = 6;
    int[] diceCurrentValues = new int[t_throws];
    for (int i = 0; i < diceCurrentValues.length; i++) diceCurrentValues[i] = 1;

    int combinations = 0;
    int matches = 0;
    for (; ; ) {
        // calculate the sum of the current combination
        int sum = 0;
        for (int diceValue : diceCurrentValues) sum += diceValue;

        if (sum % x_multiple == 0) matches++;
        System.out.println("status: " + matches + "/" + combinations + "=" + (matches * 100 / (double) combinations) + "%");

        // create the next dice combination
        int dicePointer = 0;
        boolean incremented = false;
        while (!incremented) {
            if (dicePointer == diceCurrentValues.length) return;
            if (diceCurrentValues[dicePointer] == s_sides) {
                diceCurrentValues[dicePointer] = 1;
            } else {
                incremented = true;


Here's another example. If you throw a 6-sided dice 10 times, there is 1/4 probability that the sum is a multiple of 4. The program above should run with the following parameters:

    int t_throws = 10;
    int s_sides = 6;
    int x_multiple = 4;

The program will show the final output: status: 15116544/60466176=25.0% That means that there are 60466176 combinations (i.e. 6^10) and that there are 15116544 of them where the sum is a multiple of 4. So, that's 25% (=1/4).

This just follows the formula as mentioned above (i.e. P(s,t,x) = 1/x). x is 4 in this case.

  • $\begingroup$ "Assuming that t converges to infinity. This is also the case when s<x." This sounds nonsensical. $\endgroup$
    – djechlin
    Nov 11 '15 at 17:43
  • $\begingroup$ Excuse me for my poor English :) The thing is, if you apply this for values greater than the number of sides of your dice. (e.g. multiples of 10 with a 6 sided dice.) then P = 1/x is no longer correct. But it does converge 1/x ; meaning that if you would throw an infinit amount of times, the result would be 1/x again. $\endgroup$
    – bvdb
    Nov 12 '15 at 0:10
  • $\begingroup$ I made some slight adjustments. Let me know what you think. $\endgroup$
    – bvdb
    Nov 12 '15 at 0:22
  • $\begingroup$ It's not the case that the limiting factor is if s>=x. Consider the probability that 10 rolls of a six-sided dice divides 4. (Of course the limit still converges when the number of dices increases) $\endgroup$
    – Taemyr
    Nov 12 '15 at 8:40
  • $\begingroup$ @Taemyr for 10 roles with a 6-sided dice, there are 60466176 combinations, of which 15116544 have a sum which is a multiple of 4. That's exactly 25% which is exactly 1/4. So, it's correct, right ? $\endgroup$
    – bvdb
    Nov 14 '15 at 15:51

Roll the die 9 times and add up the dots. The answer is x. Roll the die one more time. add the number thrown to x to get one and only one of the following answers; x+1, x+2, x+3, x+4, x+5 or x+6. since these answers are six sequential numbers one and only one of them will be divisible by six. Therefore the probability of the sum of ten rolls of a die being divisible by six is exactly 1/6.


Couldn't you also think of it as the maximum possible value you could get for rolling the die 10 times would be 60, how many numbers between 1 and 60 are divisible by 6? 10 numbers are divisible by 6(6*1, 2, 3, etc.) So, 10 out of 60 possible values gives... 1/6. I love math.

  • 3
    $\begingroup$ True, but some numbers appear multiple times. So, even though your answer is correct, your logic is flawed. $\endgroup$
    – bvdb
    Nov 11 '15 at 11:58
  • 6
    $\begingroup$ The minimum possible value is 10. So only 9 of 51 values are divisible by 6. But not all numbers are equally probable. $\endgroup$
    – Cephalopod
    Nov 11 '15 at 11:58
  • $\begingroup$ The numbers are not distributed evenly. $\endgroup$
    – fleablood
    Nov 11 '15 at 20:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.